1 First Order Partial Differential Equations

Transcription

1 Firs Order Parial Differenial Eqaions The profond sdy of nare is he mos ferile sorce of mahemaical discoveries. - Joseph Forier ( ). Inrodcion We begin or sdy of parial differenial eqaions wih firs order parial differenial eqaions. Before doing so, we need o define a few erms. Recall (see he appendi on differenial eqaions) ha an n-h order ordinary differenial eqaion is an eqaion for an nknown fncion y() ha epresses a relaionship beween he nknown fncion and is firs n derivaives. One cold wrie his generally as n-h order ordinary differenial eqaion F(y (n) (), y (n ) (),..., y (), y(), ) = 0. (.) Here y (n) () represens he nh derivaive of y(). Frhermore, and iniial vale problem consiss of he differenial eqaion pls he vales of he Iniial vale problem. firs n derivaives a a pariclar vale of he independen variable, say 0 : y (n ) ( 0 ) = y n, y (n 2) ( 0 ) = y n 2,..., y( 0 ) = y 0. (.2) If condiions are insead provided a more han one vale of he independen variable, hen we have a bondary vale problem.. If he nknown fncion is a fncion of several variables, hen he derivaives are parial derivaives and he resling eqaion is a parial differenial eqaion. Ths, if = (, y,...), a general parial differenial eqaion migh ake he form ( F, y,...,,, ) y,..., 2 2,... = 0. (.3) Since he noaion can ge cmbersome, here are differen ways o wrie he parial derivaives. Firs order derivaives cold be wrien as,,, D.

2 2 parial differenial eqaions Second order parial derivaives cold be wrien in he forms 2 2,,, D 2. 2 y = 2 y, y, y, D y D. Noe, we are assming ha (, y,...) has coninos parial derivaives. Then, according o Claira s Theorem (Aleis Clade Claira, ), mied parial derivaives are he same. Eamples of some of he parial differenial eqaion reaed in his book are shown in Table 2.. However, being ha he highes order derivaives in hese eqaion are of second order, hese are second order parial differenial eqaions. In his chaper we will focs on firs order parial differenial eqaions. Eamples are given by + = 0. + = 0. + =. 3 2 y + =. For fncion of wo variables, which he above are eamples, a general firs order parial differenial eqaion for = (, y) is given as F(, y,,, y ) = 0, (, y) D R 2. (.4) Linear firs order parial differenial eqaion. Qasilinear firs order parial differenial eqaion. Semilinear firs order parial differenial eqaion. This eqaion is oo general. So, resricions can be placed on he form, leading o a classificaion of firs order eqaions. A linear firs order parial differenial eqaion is of he form a(, y) + b(, y) y + c(, y) = f (, y). (.5) Noe ha all of he coefficiens are independen of and is derivaives and each erm in linear in,, or y. We can rela he condiions on he coefficiens a bi. Namely, we cold assme ha he eqaion is linear only in and y. This gives he qasilinear firs order parial differenial eqaion in he form a(, y, ) + b(, y, ) y = f (, y, ). (.6) Noe ha he -erm was absorbed by f (, y, ). In beween hese wo forms we have he semilinear firs order parial differenial eqaion in he form a(, y) + b(, y) y = f (, y, ). (.7) Here he lef side of he eqaion is linear in, and y. However, he righ hand side can be nonlinear in. For he mos par, we will inrodce he Mehod of Characerisics for solving qasilinear eqaions. B, le s firs consider he simpler case of linear firs order consan coefficien parial differenial eqaions.

3 firs order parial differenial eqaions 3.2 Linear Consan Coefficien Eqaions Le s consider he linear firs order consan coefficien parial differenial eqaion a + b y + c = f (, y), (.8) for a, b, and c consans wih a 2 + b 2 > 0. We will consider how sch eqaions migh be solved. We do his by considering wo cases, b = 0 and b = 0. For he firs case, b = 0, we have he eqaion a + c = f. We can view his as a firs order linear (ordinary) differenial eqaion wih y a parameer. Recall ha he solion of sch eqaions can be obained sing an inegraing facor. [See he discssion afer Eqaion (B.7).] Firs rewrie he eqaion as Inrodcing he inegraing facor + c a = f a. c µ() = ep( a dξ) = e a c, he differenial eqaion can be wrien as (µ) = f a µ. Inegraing his eqaion and solving for (, y), we have µ()(, y) = a e a c (, y) = a (, y) = a f (ξ, y)µ(ξ) dξ + g(y) f (ξ, y)e c a ξ dξ + g(y) f (ξ, y)e c a (ξ ) dξ + g(y)e c a. (.9) Here g(y) is an arbirary fncion of y. For he second case, b = 0, we have o solve he eqaion a + b y + c = f. I wold help if we cold find a ransformaion which wold eliminae one of he derivaive erms redcing his problem o he previos case. Tha is wha we will do. We firs noe ha a + b y = (ai + bj) ( i + y j) = (ai + bj). (.0)

4 4 parial differenial eqaions z = y ai + bj w = b ay Figre.: Coordinae sysems for ransforming a + b y + c = f ino bv z + cv = f sing he ransformaion w = b ay and z = y. Recall from mlivariable calcls ha he las erm is nohing b a direcional derivaive of (, y) in he direcion ai + bj. [Acally, i is proporional o he direcional derivaive if ai + bj is no a ni vecor.] Therefore, we seek o wrie he parial differenial eqaion as involving a derivaive in he direcion ai + bj b no in a direcional orhogonal o his. In Figre. we depic a new se of coordinaes in which he w direcion is orhogonal o ai + bj. We consider he ransformaion w = b ay, We firs noe ha his ransformaion is inverible, z = y. (.) = (w + az), b y = z. (.2) Ne we consider how he derivaive erms ransform. Le (, y) = v(w, z). Then, we have a + b y = a v(w, z) + b v(w, z), y [ v w = a w + v ] z z [ v w +b w y + v ] z z y = a[bv w + 0 v z ] + b[ av w + v z ] = bv z. (.3) Therefore, he parial differenial eqaion becomes ( ) bv z + cv = f (w + az), z. b This is now in he same form as in he firs case and can be solved sing an inegraing facor. Eample.. Find he general solion of he eqaion 3 2 y + =. Firs, we ransform he eqaion ino new coordinaes. and z = y. The, w = b ay = 2 3y, 2 y = 3[ 2v w + 0 v z ] 2[ 3v w + v z ] = 2v z. (.4) The new parial differenial eqaion for v(w, z) is 2 v z + v = = (w + 3z). 2

5 firs order parial differenial eqaions 5 Rewriing his eqaion, we idenify he inegraing facor v z 2 v = (w + 3z), 4 µ(z) = ep [ z ] 2 dζ = e z/2. Using his inegraing facor, we can solve he differenial eqaion for v(w, z). ( ) e z/2 v = z 4 (w + 3z)e z/2, e z/2 v(w, z) = z(w + 3ζ)e ζ/2 dζ 4 = 2 (w z)e z/2 + c(w) v(w, z) = (w z) + c(w)ez/2 2 (, y) = 3 + c( 2 3y)e y/2. (.5).3 Qasilinear Eqaions: The Mehod of Characerisics.3. Geomeric Inerpreaion We consider he qasilinear parial differenial eqaion in wo independen variables, a(, y, ) + b(, y, ) y c(, y, ) = 0. (.6) Le = (, y) be a solion of his eqaion. Then, f (, y, ) = (, y) = 0 describes he solion srface, or inegral srface, We recall from mlivariable, or vecor, calcls ha he normal o he inegral srface is given by he gradien fncion, Inegral srface. f = (, y, ). Now consider he vecor of coefficiens, v = (a, b, c) and he do prodc wih he gradien above: v f = a + b y c. This is he lef hand side of he parial differenial eqaion. Therefore, for he solion srface we have v f = 0, or v is perpendiclar o f. Since f is normal o he srface, v = (a, b, c) is angen o he srface. Geomerically, v defines a direcion field, called he characerisic field. These are shown in Figre.2. The characerisic field.

6 6 parial differenial eqaions.3.2 Characerisics We seek he forms of he characerisic crves sch as he one shown in Figre.2. Recall ha one can paramerize space crves, c() = ((), y(), ()), [, 2 ]. The angen o he crve is hen v() = dc() = ( d, dy, d ). However, in he las secion we saw ha v() = (a, b, c) for he parial differenial eqaion a(, y, ) + b(, y, ) y c(, y, ) = 0. This gives he parameric form of he characerisic crves as d = a, dy = b, d = c. (.7) Anoher form of hese eqaions is fond by relaing he differenials, d, dy, d, o he coefficiens in he differenial eqaion. Since = () and y = y(), we have dy d = dy/ d/ = b a. Similarly, we can show ha d d = c a, d dy = c b. All of hese relaions can be smmarized in he form = d a = dy b = d c. (.8) How do we se hese characerisics o solve qasilinear parial differenial eqaions? Consider he ne eample. Eample.2. Find he general solion: + y = 0. We firs idenify a =, b =, and c =. The relaions beween he differenials is d = dy = d. We can pair he differenials in hree ways: dy d =, d d =, d dy =. Only wo of hese relaions are independen. We focs on he firs pair. The firs eqaion gives he characerisic crves in he y-plane. This eqaion is easily solved o give y = + c. The second eqaion can be solved o give = c 2 e.

7 firs order parial differenial eqaions 7 The goal is o find he general solion o he differenial eqaion. Since = (, y), he inegraion consan is no really a consan, b is consan wih respec o. I is in fac an arbirary consan fncion. In fac, we cold view i as a fncion of c, he consan of inegraion in he firs eqaion. Ths, we le c 2 = G(c ) for G and arbirary fncion. Since c = y, we can wrie he general solion of he differenial eqaion as (, y) = G(y )e. Eample.3. Solve he advecion eqaion, + c = 0, for c a consan, and = (, ), <, > 0. The characerisic eqaions are dτ = = d c = d 0 (.9) and he parameric eqaions are given by These eqaions imply ha = cons. = c. d dτ = c, d = 0. (.20) dτ = c + cons. = c + c 2. As before, we can wrie c as an arbirary fncion of c 2. However, before doing so, le s replace c wih he variable ξ and hen we have ha Traveling waves. ξ = c, (, ) = f (ξ) = f ( c) where f is an arbirary fncion. Frhermore, we see ha (, ) = f ( c) indicaes ha he solion is a wave moving in one direcion in he shape of he iniial fncion, f (). This is known as a raveling wave. A ypical raveling wave is shown in Figre.3. Noe ha since = (, ), we have 0 = + c = + d = d((),. (.2) f () f ( c) Figre.3: Depicion of a raveling wave. (, ) = f () a = 0 ravels wiho changing shape. c This implies ha (, ) = consan along he characerisics, d = c. As wih ordinary differenial eqaions, he general solion provides an infinie nmber of solions of he differenial eqaion. If we wan o pick o a pariclar solion, we need o specify some side condiions. We Side condiions. invesigae his by way of eamples. Eample.4. Find solions of + y = 0 sbjec o (, 0) =.

8 8 parial differenial eqaions We fond he general solion o he parial differenial eqaion as (, y) = G(y )e. The side condiion ells s ha = along y = 0. This reqires = (, 0) = G( )e. Ths, G( ) = e. Replacing wih z, we find G(z) = e z. Ths, he side condiion has allowed for he deerminaion of he arbirary fncion G(y ). Insering his fncion, we have (, y) = G(y )e = e y e = e y. Side condiions cold be placed on oher crves. For he general line, y = m + d, we have (, m + d) = g() and for = d, (d, y) = g(y). As we will see, i is possible ha a given side condiion may no yield a solion. We will see ha condiions have o be given on non-characerisic crves in order o be sefl. Eample.5. Find solions of 3 2 y + = for a) (, ) = and b) (, y) = 0 on 3y + 2 =. Before applying he side condiion, we find he general solion of he parial differenial eqaion. Rewriing he differenial eqaion in sandard form, we have The characerisic eqaions are 3 2 y = =. These eqaions imply ha d 3 = dy 2 = d. (.22) 2d = 3dy This implies ha he characerisic crves (lines) are 2 + 3y = c. d d = 3 ( ). This is a linear firs order differenial eqaion, d d + 3 = 3. I can be solved sing he inegraing facor, ( ) µ() = ep dξ = e /3. 3 d ( e /3) = d 3 e/3 e /3 = ξe ξ/3 dξ + c 2 3 = ( 3)e /3 + c 2 (, y) = 3 + c 2 e /3. (.23)

9 firs order parial differenial eqaions 9 As before, we wrie c 2 as an arbirary fncion of c = 2 + 3y. This gives he general solion (, y) = 3 + G(2 + 3y)e /3. Noe ha his is he same answer ha we had fond in Eample. Now we can look a any side condiions and se hem o deermine pariclar solions by picking o specific G s. a (, ) = This saes ha = along he line y =. Insering his condiion ino he general solion, we have or Leing z = 5, = 3 + G(5)e /3, G(5) = 3e /3. G(z) = 3e z/5. The pariclar solion saisfying his side condiion is (, y) = 3 + G(2 + 3y)e /3 = 3 + 3e (2+3y)/5 e /3 = 3 + 3e (y )/5. (.24) This srface is shown in Figre.5. In Figre.5 we sperimpose he vales of (, y) along he characerisic crves. The characerisic crves are he red lines and he images of hese crves are he black lines. The side condiion is indicaed wih he ble crve drawn along he srface. The vales of (, y) are fond from he side condiion as follows. For = ξ on he ble crve, we know ha y = ξ and (ξ, ξ) = ξ. Now, he characerisic lines are given by 2 + 3y = c. The consan c is fond on he ble crve from he poin of inersecion wih one of he black characerisic lines. For = y = ξ, we have c = 5ξ. Then, he eqaion of he characerisic line, which is red in Figre.5, is given by y = 3 (5ξ 2). Along hese lines we need o find (, y) = 3 + c 2 e /3. Firs we have o find c 2. We have on he ble crve, ha ξ = (ξ, ξ) = ξ 3 + c 2 e ξ/3. (.25) Therefore, c 2 = 3e ξ/3. Insering his resl ino he epression for he solion, we have (, y) = 3 + e (ξ )/3. So, for each ξ, one can draw a family of spacecrves (, 3 (5ξ 2), 3 + e(ξ )/3 ) Figre.4: Inegral srface fond in Eample.5. Figre.5: Inegral srface wih side condiion and characerisics for Eample.5. yielding he inegral srface.

10 0 parial differenial eqaions b (, y) = 0 on 3y + 2 =. For his condiion, we have 0 = 3 + G()e /3. We noe ha G is no a fncion in his epression. We only have one vale for G. So, we canno solve for G(). Geomerically, his side condiion corresponds o one of he black crves in Figre.5..4 Applicaions.4. Conservaion Laws There are many applicaions of qasilinear eqaions, especially in flid dynamics. The advecion eqaion is one sch eample and generalizaions of his eample o nonlinear eqaions leads o some ineresing problems. These eqaions fall ino a caegory of eqaions called conservaion laws. We will firs discss one-dimensional (in space) conservaions laws and hen look a simple eamples of nonlinear conservaion laws. Conservaion laws are sefl in modeling several sysems. They can be boiled down o deermining he rae of change of some sff, Q(), in a region, a b, as depiced in Figre.6. The simples model is o hink of flid flowing in one dimension, sch as waer flowing in a sream. Or, i cold be he ranspor of mass, sch as a pollan. One cold hink of raffic flow down a sraigh road. Figre.6: The rae of change of Q beween = a and = b depends on he raes of flow hrogh each end. φ(a, ) Q() φ(b, ) = a = b This is an eample of a ypical miing problem. The rae of change of Q() is given as he rae of change of Q = Rae in Rae O + sorce erm. Here he Rae in is how mch is flowing ino he region in Figre.6 from he = a bondary. Similarly, he Rae o is how mch is flowing ino he region from he = b bondary. [Of corse, his cold be he oher way, b we can imagine for now ha q is flowing from lef o righ.] We can describe his flow in erms of he fl, φ(, ) over he ends of he region. On he lef side we have a gain of φ(a, ) and on he righ side of he region here is a loss of φ(b, ). The sorce erm wold be some oher means of adding or removing Q from he region. In erms of flid flow, here cold be a sorce of flid

11 firs order parial differenial eqaions inside he region sch as a face adding more waer. Or, here cold be a drain leing waer escape. We can denoe his by he oal sorce over he inerval, b a f (, ) d. Here f (, ) is he sorce densiy. In smmary, he rae of change of Q(, ) can be wrien as dq = φ(a, ) φ(b, ) + b a f (, y) d. We can wrie his in a slighly differen form by noing ha φ(a, ) φ(b, ) can be viewed as he evalaion of aniderivaives in he Fndamenal Theorem of Calcls. Namely, we can recall ha b a φ(, ) d = φ(b, ) φ(a, ). The difference is no eacly in he order ha we desire, b i is easy o see ha dq b = φ(, ) b d + f (, ) d. (.26) a a This is he inegral form of he conservaion law. We can rewrie he conservaion law in differenial form. Firs, we inrodce he densiy fncion, (, ), so ha he oal amon of sff a a given ime is Q() = b a (, ) d. Inrodcing his form ino he inegral conservaion law, we have d b b φ b (, ) d = a a d + f (, ) d. (.27) a Assming ha a and b are fied in ime and ha he inegrand is coninos, we can bring he ime derivaive inside he inegrand and collec he hree erms ino one o find b ( (, ) + φ (, ) f (, )) d = 0, [a, b]. a We canno simply se he inegran o zero js becase he inegral vanishes. However, if his resl holds for every region [a, b], hen we can conclde he inegrand vanishes. So, nder ha assmpion, we have he local conservaion law, Inegral form of conservaion law. Differenial form of conservaion law. (, ) + φ (, ) = f (, ). (.28) This parial differenial eqaion is acally an eqaion in erms of wo nknown fncions, assming we know somehing abo he sorce fncion. We wold like o have a single nknown fncion. So, we need some addiional informaion. This added informaion comes from he consiive relaion, a fncion relaing he fl o he densiy fncion. Namely, we will assme ha we can find he relaionship φ = φ(). If so, hen we can wrie or φ = φ (). φ = dφ d,

12 2 parial differenial eqaions Eample.6. Inviscid Brgers Eqaion Find he eqaion saisfied by (, ) for φ() = 2 2 and f (, ) 0. For his fl fncion we have φ = φ () =. The resling eqaion is hen + = 0. This is he inviscid Brgers eqaion. We will laer discss Brgers eqaion. v v Figre.7: Car velociy as a fncion of car densiy. Eample.7. Traffic Flow This is a simple model of one-dimensional raffic flow. Le (, ) be he densiy of cars. Assme ha here is no sorce erm. For eample, here is no way for a car o disappear from he flow by rning off he road or falling ino a sinkhole. Also, here is no sorce of addiional cars. Le φ(, ) denoe he nmber of cars per hor passing posiion a ime. Noe ha he nis are given by cars/mi imes mi/hr. Ths, we can wrie he fl as φ = v, where v is he velociy of he cars a posiion and ime. In order o conine we need o assme a relaionship beween he car velociy and he car densiy. Le s assme he simples form, a linear relaionship. The more dense he raffic, we epec he speeds o slow down. So, a fncion similar o ha in Figre.7 is in order. This is a sraigh line beween he wo inerceps (0, v ) and (, 0). I is easy o deermine he eqaion of his line. Namely he relaionship is given as v = v v. This gives he fl as ( ) φ = v = v 2. We can now wrie he eqaion for he car densiy, 0 = + φ ( = + v 2 ). (.29).4.2 Nonlinear Advecion Eqaions In his secion we consider eqaions of he form + c() = 0. When c() is a consan fncion, we have he advecion eqaion. In he las wo eamples we have seen cases in which c() is no a consan fncion. We will apply he mehod of characerisics o hese eqaions. Firs, we will recall how he mehod works for he advecion eqaion. The advecion eqaion is given by + c = 0. The characerisic eqaions are given by d = c, d = 0. These are easily solved o give he resl ha (, ) = consan along he lines = c + 0, where 0 is an arbirary consan.

13 firs order parial differenial eqaions 3 The characerisic lines are shown in Figre.8. We noe ha (, ) = ( 0, 0) = f ( 0 ). So, if we know iniially, we can deermine wha is a a laer ime. Figre.8: The characerisics lines he -plane. = slope = /c 0 ( 0 + c, ) = ( 0, 0) In Figre.8 we see ha he vale of ( 0, ) a = 0 and = 0 propagaes along he characerisic o a poin a ime =. From c = 0, we can solve for in erms of and find ha ( 0 + c, ) = ( 0, 0). Plos of solions (, ) verss for specific imes give raveling waves as shown in Figre.3. In Figre.9 we show how each wave profile for differen imes are consrced for a given iniial condiion. The nonlinear advecion eqaion is given by + c() = 0, <. Le (, 0) = 0 () be he iniial profile. The characerisic eqaions are given by d = c(), These are solved o give he resl ha d = 0. 0 Figre.9: For each = 0 a = 0, ( 0 + c, ) = ( 0, 0). (, ) = consan, along he characerisic crves () = c(). The lines passing hogh ( 0, ) = 0 ( 0 ) have slope /c( 0 ( 0 )). Eample.8. Solve + = 0, (, 0) = e 2. For his problem = consan along d =. Since is consan, his eqaion can be inegraed o yield = ( 0, 0) + 0. Insering he iniial condiion, = e Therefore, he solion is (, ) = e 2 0 along = e In Figre.0 he characerisics a shown. In his case we see ha he characerisics inersec. In Figre charlines3 we look more specifically a he inersecion of he characerisic lines for 0 = 0 and 0 =. These are approimaely he firs lines o inersec; i.e., here are (almos) no inersecions a earlier imes. A he

14 4 parial differenial eqaions Figre.0: The characerisics lines he -plane for he nonlinear advecion eqaion. slope = e 2 0 Figre.: The characerisics lines for 0 = 0, in he -plane for he nonlinear advecion eqaion. = = e 0 = 0 0 = =0.0 =0.5 =.0 inersecion poin he fncion (, ) appears o ake on more han one vale. For he case shown, he solion wans o ake he vales = 0 and =. In Figre.2 we see he developmen of he solion. This is fond sing a parameric plo of he poins ( 0 + e 2 0, e 2 0) for differen imes. The iniial profile propagaes o he righ wih he higher poins raveling faser han he lower poins since () = > 0. Arond =.0 he wave breaks and becomes mlivaled. The ime a which he fncion becomes mlivaled is called he breaking ime. = The Breaking Time =2.0 Figre.2: The developmen of a gradien caasrophe in Eample.8 leading o a mlivaled fncion. In he las eample we saw ha for nonlinear wave speeds a gradien caasrophe migh occr. The firs ime a which a caasrophe occrs is called he breaking ime. We will deermine he breaking ime for he nonlinear advecion eqaion, + c() = 0. For he characerisic corresponding o 0 = ξ, he wavespeed is given by and he characerisic line is given by F(ξ) = c( 0 (ξ)) = ξ + F(ξ). 0 (ξ) = (ξ, 0). The vale of he wave fncion along his characerisic is (, ) = (ξ + F(ξ), ) =. (.30) Therefore, he solion is (, ) = 0 (ξ) along = ξ + F(ξ).

15 firs order parial differenial eqaions 5 This means ha = 0 (ξ)ξ and = 0 (ξ)ξ. We can deermine ξ and ξ sing he characerisic line Then, we have ξ = F(ξ). ξ = F (ξ)ξ = + F (ξ). ξ = ( F(ξ)) = F(ξ) F (ξ)ξ = F(ξ) + F (ξ). (.3) Noe ha ξ and ξ are ndefined if he denominaor in boh epressions vanishes, + F (ξ) = 0, or a ime = F (ξ). The minimm ime for his o happen in he breaking ime, { b = min } F. (.32) (ξ) The breaking ime. Eample.9. Find he breaking ime for + = 0, (, 0) = e 2. Since c() =, we have and This gives F(ξ) = c( 0 (ξ)) = e ξ2 F (ξ) = 2ξe ξ2. = 2ξe ξ2. We need o find he minimm ime. Ths, we se he derivaive eqal o zero and solve for ξ. ( ) 0 = d e ξ2 dξ 2ξ = (2 ξ ) e ξ (.33) Ths, he minimm occrs for 2 ξ 2 ( ) b = 2 = = 0, or ξ = / 2. This gives 2 2e /2 = e.6. (.34) 2

16 6 parial differenial eqaions.4.4 Shock Waves Weak solions. Solions of nonlinear advecion eqaions can become mlivaled de o a gradien caasrophe. Namely, he derivaives and become ndefined. We wold like o eend solions pas he caasrophe. However, his leads o he possibiliy of disconinos solions. Sch solions which may no be differeniable or coninos in he domain are known as weak solions. In pariclar, consider he iniial vale problem + φ = 0, R, > 0, (, 0) = 0 (). =.5 =.75 =2.0 Figre.3: The shock solion afer he breaking ime. + s s Figre.4: Depicion of he jmp disconiniy a he shock posiion. R R + Figre.5: Domains on eiher side of shock pah are denoed as R + and R. a b Then, (, ) is a weak solion of his problem if 0 [v + φv ] d + 0 ()v(, 0) d = 0 for all smooh fncions v C (R [0, )) wih compac sppor, i.e., v = 0 oside some compac sbse of he domain. Effecively, he weak solion ha evolves will be a piecewise smooh fncion wih a disconiniy, he shock wave, ha propagaes wih shock speed. I can be shown ha he form of he shock will be he disconiniy shown in Figre.3 sch ha he areas c from he solions will cancel leaving he oal area nder he solion consan. [See G. B. Whiham s Linear and Nonlinear Waves, 973.] We will consider he disconiniy as shown in Figre.4. We can find he eqaion for he shock pah by sing he inegral form of he conservaion law, d b (, ) d = φ(a, ) φ(b, ). a Recall ha one can differeniae nder he inegral if (, ) and (, ) are coninos in and in an appropriae sbse of he domain. In pariclar, we will inegrae over he inerval [a, b] as shown in Figre.5. The domains on eiher side of shock pah are denoed as R + and R and he limis of () and (, ) as one approaches from he lef of he shock are denoed by s () and = (s, ). Similarly, he limis of () and (, ) as one approaches from he righ of he shock are denoed by s + () and + = ( s +, ). We need o be carefl in differeniaing nder he inegral, [ d b (, ) d = d ] s () b (, ) d + (, ) d a a s + () s () b = (, ) d + (, ) d a s + () +(s, ) d s ( s +, ) d+ s = φ(a, ) φ(b, ). (.35)

17 firs order parial differenial eqaions 7 Taking he limis a s and b + s, we have ha ( ( s, ) ( s +, ) ) d s = φ(s, ) φ( s +, ). Adoping he noaion [ f ] = f ( s + ) f (s ), we arrive a he Rankine-Hgonoi jmp condiion d s = [φ] []. (.36) This gives he eqaion for he shock pah as will be shown in he ne eample. Eample.0. Consider he problem + = 0, <, > 0 saisfying he iniial condiion {, 0, (, 0) = 0, > 0. The characerisics for his parial differenial eqaion are familiar by now. The iniial condiion and characerisics are shown in Figre.6. From () =, here are wo possibiliies. If = 0, hen we have a consan. If = along he characerisics, he we have sraigh lines of slope one. Therefore, he characerisics are given by () = { 0, > 0, + 0, < 0. As seen in Figre.6 he characerisics inersec immediaely a = 0. The shock pah is fond from he Rankine-Hgonoi jmp condiion. We firs noe ha φ() = 2 2, since φ =. Then, we have d s = [φ] [] = = ( + + )( + ) 2 + = 2 (+ + ) The Rankine-Hgonoi jmp condiion. = = 0 Figre.6: Iniial condiion and characerisics for Eample.0. = 2 (0 + ) = 2. (.37) Now we need only solve he ordinary differenial eqaion s() = 2 wih iniial condiion s (0) = 0. This gives s () = 2. This line separaes he characerisics on he lef and righ side of he shock solion. The solion is given by {, /2, (, ) = 0, > /2. In Figre.7 we show he characerisic lines ending a he shock pah (in red) wih = 0 and on he righ and = on he lef of he shock pah. This is consisen wih he solion. One js sees he iniial sep fncion moving o he righ wih speed /2 wiho changing shape. = = 0 Figre.7: The characerisic lines end a he shock pah (in red). On he lef = and on he righ = 0.

18 8 parial differenial eqaions.4.5 Rarefacion Waves Shocks are no he only ype of solions enconered when he velociy is a fncion of. There may someimes be regions where he characerisic lines do no appear. A simple eample is he following. Eample.. Draw he characerisics for he problem + = 0, <, > 0 saisfying he iniial condiion { 0, 0, (, 0) =, > 0. = 0 = Figre.8: Iniial condiion and characerisics for Eample.4. In his case he solion is zero for negaive vales of and posiive for posiive vales of as shown in Figre.8. Since he wavespeed is given by, he = iniial vales have he waves on he righ moving o he righ and he vales on he lef say fied. This leads o he characerisics in Figre.8 showing a region in he -plane ha has no characerisics. In his secion we will discover how o fill in he missing characerisics and, hs, he deails abo he solion beween he = 0 and = vales. As moivaion, we consider a smoohed o version of his problem. Eample.2. Draw he characerisics for he iniial condiion 0, ɛ, (, 0) = +ɛ 2ɛ, ɛ,, > ɛ. -ɛ ɛ The fncion is shown in he op graph in Figre.9. The lefmos and righmos characerisics are he same as he previos eample. The only new par is deermining he eqaions of he characerisics for ɛ. These are fond sing he mehod of characerisics as = ξ + 0 (ξ), 0 (ξ) = ξ + ɛ 2ɛ. These characerisics are drawn in Figre.9 in red. Noe ha hese lines ake on slopes varying from infinie slope o slope one, corresponding o speeds going from zero o one. -ɛ = 0 ɛ = Figre.9: The fncion and characerisics for he smoohed sep fncion. Characerisics for rarefacion, or epansion, waves are fan-like characerisics. Comparing he las wo eamples, we see ha as ɛ approaches zero, he las eample converges o he previos eample. The characerisics in he region where here were none become a fan. We can see his as follows. Since ξ < ɛ for he fan region, as ɛ ges small, so does his inerval. Le s scale ξ as ξ = σɛ, σ [, ]. Then, = σɛ + 0 (σɛ), 0 (σɛ) = σɛ + ɛ = (σ + ). 2ɛ 2 For each σ [, ] here is a characerisic. Leing ɛ 0, we have = c, c = (σ + ). 2

19 firs order parial differenial eqaions 9 Ths, we have a family of sraigh characerisic lines in he -plane passing hrogh (0, 0) of he form = c for c varying from c = 0 o c =. These are shown as he red lines in Figre.20. The fan characerisics can be wrien as / = consan. So, we can seek o deermine hese characerisics analyically and in a sraigh forward manner by seeking solions of he form (, ) = g( ). Eample.3. Deermine solions of he form (, ) = g( ) o + = 0. Insering his gess ino he differenial eqaion, we have 0 = + = g ( g ). (.38) Ths, eiher g = 0 or g =. The firs case will no work since his gives consan solions. The second solion is eacly wha we had obained before. Recall ha solions along characerisics give (, ) = = consan. The characerisics and solions for = 0,, 2 are shown in Figre rarefacionfig4. A a specific ime one can draw a line (dashed lines in figre) and follow he characerisics back o he = 0 vales, (ξ, 0) in order o consrc (, ). = 0 = Figre.20: The characerisics for Eample.4 showing he fan characerisics. Seek rarefacion fan waves sing (, ) = g( ). = 0 = 0 = = 2 = Figre.2: The characerisics and solions for = 0,, 2 for Eample.4 = = 2 As a las eample, le s invesigae a nonlinear model which possesses boh shock and rarefacion waves. Eample.4. Solve he iniial vale problem + 2 = 0, <, > 0 saisfying he iniial condiion 0, 0, (, 0) =, 0 < < 2, 0, 2.

20 20 parial differenial eqaions The mehod of characerisics gives Therefore, d = 2, d = 0. (, ) = 0 (ξ) = cons. along he lines () = 2 0 (ξ) + ξ. There are hree vales of 0 (ξ), 0, ξ 0, 0 (ξ) =, 0 < ξ < 2, 0, ξ 2. In Figre.22 we see ha here is a rarefacion and a gradien caasrophe. Figre.22: In his eample here occrs a rarefacion and a gradien caasrophe. 0 2 = 0 = = 0 In order o fill in he fan characerisics, we need o find solions (, ) = g(/). Insering his gess ino he differenial eqaion, we have 0 = + 2 = g ( g 2 ). (.39) Ths, eiher g = 0 or g 2 =. The firs case will no work since his gives consan solions. The second solion gives ( ) g =.. Therefore, along he fan characerisics he solions are (, ) = = consan. These fan characerisics are added in Figre.23. Ne, we rn o he shock pah. We see ha he firs inersecion occrs a he poin (, ) = (2, 0). The Rankine-Hgonoi condiion gives d s = [φ] [] = = ( + )( ) +

21 firs order parial differenial eqaions 2 Figre.23: The fan characerisics are added o he oher characerisic lines. = 0 = = 0 = 3 ( ) = 3 ( ) = 3. (.40) Ths, he shock pah is given by s() = 3 wih iniial condiion s(0) = 2. This gives s () = In Figre.24 he shock pah is shown in red wih he fan characerisics and verical lines meeing he pah. Noe ha he fan lines and verical lines cross he shock pah. This leads o a change in he shock pah. Figre.24: The shock pah is shown in red wih he fan characerisics and verical lines meeing he pah. = 0 = = 0 The new pah is fond sing he Rankine-Hgonoi condiion wih + = 0 and =. Ths, d s = [φ] [] = = ( + )( ) + = 3 ( ) = 3 ( s ) = s 3. (.4) We need o solve he iniial vale problem d s = 3 s, s(3) = 3. This can be done sing separaion of variables. Namely, ds =. s 3

22 22 parial differenial eqaions This gives he solion s = 3 + c. Since he second shock solion sars a he poin (3, 3), we can deermine c = This gives he shock pah as s () = ( ) In Figre.25 we show his shock pah and he oher characerisics ending on he pah. Figre.25: The second shock pah is shown in red wih he characerisics shown in all regions. = 0 = = 0 I is ineresing o consrc he solion a differen imes based on he characerisics. For a given ime,, one draws a horizonal line in he -plane and reads off he vales of (, ) sing he vales a = 0 and he rarefacion solions. This is shown in Figre.26. The righ disconiniy in he iniial profile conines as a shock fron nil = 3. A ha ime he back rarefacion wave has cagh p o he shock. Afer = 3, he shock propagaes forward slighly slower and he heigh of he shock begins o decrease. De o he fac ha he parial differenial eqaion is a conservaion law, he area nder he shock remains consan as i sreches and decays in amplide..4.6 Traffic Flow An ineresing applicaion is ha of raffic flow. We had already derived he fl fncion. Le s invesigae eamples wih varying iniial condiions ha lead o shock or rarefacion waves. As we had seen earlier in modeling raffic flow, we can consider he fl fncion which leads o he conservaion law ( ) φ = v = v 2, + v ( 2 ) = 0. Here (, ) represens he densiy of he raffic and is he maimm densiy and v is he iniial velociy.

23 firs order parial differenial eqaions 23 = 0 = = 0 = 0 = 5 = 4 = 3 = 2 = Figre.26: Solions for he shockrarefacion eample. = 0 2 = = = =

24 24 parial differenial eqaions Firs, consider he flow of raffic vas i approaches a red ligh as shown in Figre.27. The raffic ha is sopped has reached he maimm densiy. The incoming raffic has a lower densiy, 0. For his red ligh problem, we consider he iniial condiion (, 0) = { 0, < 0,, 0. Figre.27: ligh. Cars approaching a red 0 < cars/mi cars/mi The characerisics for his problem are given by = c(( 0, )) + 0, 0 where c(( 0, )) = v ( 2( 0, 0) ). Since he iniial condiion is a piecewise-defined fncion, we need o consider wo cases. Firs, for 0, we have 0 Figre.28: Iniial condiion and characerisics for he red ligh problem. 0 0 Figre.29: The addiion of he shock pah for he red ligh problem. c(( 0, )) = c( ) = v ( 2 ) = v. Therefore, he slopes of he characerisics, = v + 0 are /v. For 0 < 0, we have c(( 0, )) = c( 0 ) = v ( 2 0 ). So, he characerisics are = v ( 2 0 ) + 0. In Figre.28 we plo he iniial condiion and he characerisics for < 0 and > 0. We see ha here are crossing characerisics and he begin crossing a = 0. Therefore, he breaking ime is b = 0. We need o find he shock pah saisfying s (0) = 0. The Rankine-Hgonoi condiions give d s = [φ] [] = = v = v 0. (.42) Ths, he shock pah is fond as s () = v 0.

25 firs order parial differenial eqaions 25 In Figre.29 we show he shock pah. In he op figre he red line shows he pah. In he lower figre he characerisics are sopped on he shock pah o give he complee picre of he characerisics. The picre was drawn wih v = 2 and 0 / = /3. The ne problem o consider is sopped raffic as he ligh rns green. The cars in Figre.30 begin o fan o when he raffic ligh rns green. In his model he iniial condiion is given by (, 0) = {, 0, 0, > 0. Figre.30: Cars begin o fan o when he raffic ligh rns green. cars/mi 0 cars/mi Again, c(( 0, )) = v ( 2( 0, 0) ). Insering he iniial vales of ino his epression, we obain consan speeds, ±v. The resling characerisics are given by { v + () = 0, 0, v + 0, > 0. This leads o a rarefacion wave wih he solion in he rarefacion region given by (, ) = g(/) = ( 2 ). v The characerisics are shown in Figre??. The fll solion is hen, v, (, ) = g(/), < v, 0, > v. Figre.3: The characerisics for he green ligh problem. 0.5 General Firs Order PDEs We have spen ime solving qasilinear firs order parial differenial eqaions. We now rn o nonlinear firs order eqaions of he form F(, y,,, y ) = 0,

26 26 parial differenial eqaions for = (, y). If we inrodce new variables, p = and q = y, hen he differenial eqaion akes he form F(, y,, p, q) = 0. Noe ha for (, ) a fncion wih coninos derivaives, we have p y = y = y = q. We can view F = 0 as a srface in a five dimensional space. Since he argmens are fncions of and y, we have from he mlivariable Chain Rle ha df = F + F d + F p p + F q q 0 = F + pf + p F p + p y F q. (.43) This can be rewrien as a qasilinear eqaion for p(, y) : The characerisic eqaions are Similarly, from df dy F p p + F q p = F pf. d = dy = dp. F p F q F + pf = 0 we have ha Frhermore, since = (, y), d = dy = dq. F p F q F y + qf d = d + y dy = pd + qdy = pd + q F q d F p ( = p + q F ) q. (.44) F p The Charpi eqaions. These were named afer he French mahemaician Pal Charpi Villecor, who was probably he firs o presen he mehod in his hesis he year of his deah, 784. His work was frher eended in 797 by Lagrange and given a geomeric eplanaion by Gaspard Monge (746-88) in 808. This mehod is ofen called he Lagrange-Charpi mehod. Therefore, d F p = d pf p + qf q. Combining hese resls we have he Charpi Eqaions d F p = dy F q = d = dp = dq. (.45) pf p + qf q F + pf F y + qf These eqaions can be sed o find solions of nonlinear firs order parial differenial eqaions as seen in he following eamples.

27 firs order parial differenial eqaions 27 Eample.5. Find he general solio of 2 + y y = 0. Firs, we inrodce = p and y = q. Then, Ne we idenify Then, F(, y,, p, q) = p 2 + qy = 0. F p = 2p, F q = y, F =, F = 0,, F y = q. The Charpi eqaions are hen pf p + qf q = 2p 2 + qy, F + pf = p, F y + qf = q q = 0. d 2p = dy y = d 2p 2 + qy = dp p = dq 0. The firs conclsion is ha q = c = consan. So, from he parial differenial eqaion we have = p 2 + c y. Since d = pd + qdy = pd + c dy, hen d cdy = c y d. Therefore, d( c y) c y = d z z = + c 2 2 c y = + c 2. (.46) Solving for, we have (, y) = 4 ( + c 2) 2 + c y. This eample reqired a few ricks o implemen he solion. Someimes one needs o find parameric solions. Also, if an iniial condiion is given, one needs o find he pariclar solion. In he ne eample we show how parameric solions are fond o he iniial vale problem. Eample.6. Solve he iniial vale problem 2 + y + = 0, (, 0) =. We consider he parameric form of he Charpi eqaions, = d F p = dy F q = This leads o he sysem of eqaions d = dp = dq. (.47) pf p + qf q F + pf F y + qf d = F p = 2p.

28 28 parial differenial eqaions dy d dp dq = F q =. = pf p + qf q = 2p 2 + q. = (F + pf ) = p. = (F y + qf ) = q. The second, forh, and fifh eqaions can be solved o obain y = + c. p = c 2 e. q = c 3 e. Insering hese resls ino he remaining eqaions, we have d d = 2c 2 e. = 2c 2 2 e 2 + c 3 e. These eqaions can be inegraed o find Insering hese resls ino he remaining eqaions, we have = 2c 2 e + c 4. = c 2 2 e 2 c 3 e + c 5. This is a parameric se of eqaions for (, ). Since we have e = c 4 2c 2, (, y) = c 2 2 e 2 c 3 e + c 5. ( ) 2 ( ) = c 2 c4 c4 2 c 3 + c 5 2c 2 2c 2 = 4 ( c 4) 2 + c 3 2c 2 ( c 4 ). (.48) We can se he iniial condiions by firs paramerizing he condiions. Le (s, 0) = s and y(s, 0) = 0, Then, (s, 0) = s. Since (, 0) =, (, 0) =, or p(s, 0) =. From he parial differenial eqaion, we have p 2 + q + = 0. Therefore, These relaions imply ha q(s, 0) = p 2 (s, 0) (s, 0) = ( + s). y(s, ) 0 = 0 c = 0. p(s, ) 0 = c 2 =. q(s, ) 0 = ( + s) = c 3.

29 firs order parial differenial eqaions 29 So, y(s, ) =. p(s, ) = e. q(s, ) = ( + s)e. The condiions on and give (s, ) = (s + 2) 2e, (s, ) = (s + )e e 2..6 Modern Nonlinear PDEs The sdy of nonlinear parial differenial eqaions is a ho research opic. We will (evenally) describe some eamples of imporan evolion eqaions and discss heir solions in he las chaper. Problems. Wrie he following eqaions in conservaion law form, + φ = 0 by finding he fl fncion φ(). a. + c = 0. b. + µ = 0. c = 0. d = Consider he Klein-Gordon eqaion, a = b for a and b consans. Find raveling wave solions (, ) = f ( c). 3. Find he general solion (, y) o he following problems. a. = 0. b. y y = 0. c y =. d. + y =. 4. Solve he following problems. a. + 2 y = 0, (, 0) = sin. b. + 4 = 0, (, 0) = + 2. c. y y = 0, (, 0) =. d. + = 0, (, 0) = sin. e. y + y = 0, (0, y) = e y2. f. 2 = 2, (, 0) = 2.

30 30 parial differenial eqaions g. (y ) + ( ) y = y, = 0 on y =. h. y + y = y,, y > 0, for (, 0) = e 2, > 0 and (0, y) = e y2, y > Consider he problem + = 0, <, > 0 saisfying he iniial condiion (, 0) = + 2. a. Find and plo he characerisics. b. Graphically locae where a gradien caasrophe migh occr. Esimae from yor plo he breaking ime. c. Analyically deermine he breaking ime. d. Plo solions (, ) a imes before and afer he breaking ime. 6. Consider he problem + 2 = 0, <, > 0 saisfying he iniial condiion (, 0) = + 2. a. Find and plo he characerisics. b. Graphically locae where a gradien caasrophe migh occr. Esimae from yor plo he breaking ime. c. Analyically deermine he breaking ime. d. Plo solions (, ) a imes before and afer he breaking ime. 7. Consider he problem + = 0, <, > 0 saisfying he iniial condiion { 2, 0, (, 0) =, > 0. a. Find and plo he characerisics. b. Graphically locae where a gradien caasrophe migh occr. Esimae from yor plo he breaking ime. c. Analyically deermine he breaking ime. d. Find he shock wave solion. 8. Consider he problem + = 0, <, > 0 saisfying he iniial condiion {, 0, (, 0) = 2, > 0. a. Find and plo he characerisics. b. Graphically locae where a gradien caasrophe migh occr. Esimae from yor plo he breaking ime. c. Analyically deermine he breaking ime. d. Find he shock wave solion. 9. Consider he problem + = 0, <, > 0 saisfying he iniial condiion 0,, (, 0) = 2, <,, >.

31 firs order parial differenial eqaions 3 a. Find and plo he characerisics. b. Graphically locae where a gradien caasrophe migh occr. Esimae from yor plo he breaking ime. c. Analyically deermine he breaking ime. d. Find he shock wave solion. 0. Solve he problem + = 0, <, > 0 saisfying he iniial condiion, 0, (, 0) = a, 0 < < a, 0, a.. Solve he problem + = 0, <, > 0 saisfying he iniial condiion 0, 0, (, 0) = a, 0 < < a,, a. 2. Consider he problem + 2 = 0, <, > 0 saisfying he iniial condiion { 2, 0, (, 0) =, > 0. a. Find and plo he characerisics. b. Graphically locae where a gradien caasrophe migh occr. Esimae from yor plo he breaking ime. c. Analyically deermine he breaking ime. d. Find he shock wave solion. 3. Consider he problem + 2 = 0, <, > 0 saisfying he iniial condiion {, 0, (, 0) = 2, > 0. a. Find and plo he characerisics. b. Find and plo he fan characerisics. c. Wrie o he rarefacion wave solion for all regions of he plane. 4. Solve he iniial-vale problem + = 0 <, > 0 saisfying, 0, (, 0) =, 0, 0,. 5. Consider he sopped raffic problem in a siaion where he maimm car densiy is 200 cars per mile and he maimm speed is 50 miles per hor. Assme ha he cars are arriving a 30 miles per hor. Find he solion of his problem and deermine he rae a which he raffic is backing p. How does he answer change if he cars were arriving a 5 miles per hor.

32 32 parial differenial eqaions 6. Solve he following nonlinear eqaions where p = and q = y. a. p 2 + q 2 =, (, ) =. b. pq =, (0, y) = y 2. c. p + q = pq, (, 0) =. d. pq = 2 e. p 2 + qy =. 7. Find he solion of p + qy p 2 q = 0 in parameric form for he iniial condiions a = 0 :. (, s) = s, y(, s) = 2, (, s) = s +